What is the radius of a steel sphere that wil float on water with exactly half the sphere submerged ? Density of steel is `7.9 xx 10^(3)kg//m^(3)` and surface tension of water is `7 xx 10^(-2)N`.

Options

A. 2.6 cm
B. 4.6 mm
C. 1.2 mm
D. 6.5 mm

Answer

To find the radius of a steel sphere that will float on water with exactly half the sphere submerged, we can follow these steps:

### Step 1: Understand the Forces Acting on the Sphere
When the sphere is floating with half of it submerged, there are three main forces acting on it:
1. The weight of the sphere (downward).
2. The buoyant force from the water (upward).
3. The surface tension force acting around the perimeter of the sphere (upward).

### Step 2: Set Up the Equation for Equilibrium
For the sphere to float in equilibrium, the weight of the sphere must equal the sum of the buoyant force and the surface tension force. Mathematically, this can be expressed as:
\[ \text{Weight} = \text{Buoyant Force} + \text{Surface Tension Force} \]

### Step 3: Calculate the Weight of the Sphere
The weight of the sphere can be calculated using the formula:
\[ \text{Weight} = \text{Density of Steel} \times g \times \text{Volume of Sphere} \]
The volume of a sphere is given by:
\[ \text{Volume} = \frac{4}{3} \pi R^3 \]
Thus, the weight becomes:
\[ W = \rho_s \cdot g \cdot \frac{4}{3} \pi R^3 \]
Where:
- \(\rho_s = 7.9 \times 10^3 \, \text{kg/m}^3\) (density of steel)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)

### Step 4: Calculate the Buoyant Force
The buoyant force is given by Archimedes' principle:
\[ \text{Buoyant Force} = \text{Density of Water} \times g \times \text{Volume Submerged} \]
Since half the sphere is submerged, the volume submerged is:
\[ \text{Volume Submerged} = \frac{1}{2} \cdot \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 \]
Thus, the buoyant force becomes:
\[ F_b = \rho_w \cdot g \cdot \frac{2}{3} \pi R^3 \]
Where:
- \(\rho_w = 10^3 \, \text{kg/m}^3\) (density of water)

### Step 5: Calculate the Surface Tension Force
The surface tension force acting on the sphere can be calculated as:
\[ F_t = 2 \pi R \cdot \sigma \]
Where:
- \(\sigma = 7 \times 10^{-2} \, \text{N/m}\) (surface tension of water)

### Step 6: Set Up the Equation
Now we can set up the equation for equilibrium:
\[ \rho_s \cdot g \cdot \frac{4}{3} \pi R^3 = \rho_w \cdot g \cdot \frac{2}{3} \pi R^3 + 2 \pi R \cdot \sigma \]

### Step 7: Simplify the Equation
Cancel out common terms:
\[ \rho_s \cdot \frac{4}{3} R^3 = \rho_w \cdot \frac{2}{3} R^3 + 2R \cdot \sigma \]
Rearranging gives:
\[ \frac{4}{3} \rho_s R^3 - \frac{2}{3} \rho_w R^3 = 2R \cdot \sigma \]

### Step 8: Factor Out R
Factoring out \(R\) from the left side:
\[ R^2 \left( \frac{4}{3} \rho_s - \frac{2}{3} \rho_w \right) = 2 \sigma \]

### Step 9: Solve for R
Rearranging gives:
\[ R^2 = \frac{2 \sigma}{\frac{4}{3} \rho_s - \frac{2}{3} \rho_w} \]
Substituting the values:
\[ R^2 = \frac{2 \cdot 7 \times 10^{-2}}{\frac{4}{3} \cdot 7.9 \times 10^3 - \frac{2}{3} \cdot 10^3} \]

### Step 10: Calculate R
Calculating the numerator:
\[ 2 \cdot 7 \times 10^{-2} = 0.14 \]
Calculating the denominator:
\[ \frac{4}{3} \cdot 7.9 \times 10^3 - \frac{2}{3} \cdot 10^3 = \frac{4 \cdot 7.9 - 2}{3} \times 10^3 = \frac{31.6 - 2}{3} \times 10^3 = \frac{29.6}{3} \times 10^3 \approx 9.867 \times 10^3 \]
Thus:
\[ R^2 = \frac{0.14}{9.867 \times 10^3} \approx 1.42 \times 10^{-5} \]
Taking the square root gives:
\[ R \approx 0.00377 \, \text{m} \approx 3.77 \, \text{mm} \]

### Final Answer
The radius of the steel sphere that will float on water with exactly half the sphere submerged is approximately **3.77 mm**.